Answer to Question #207907 in Mechanics | Relativity for karen

Question

From the data gathered by BSMarE student about the relative density of paraffin, paraffin has a 68.3 mass in air, mass of sinker in water is 122.4 and mass of paraffin and sinker in water is 126.5. Compute the following Add the mass of the paraffin in air to the mass of the sinker in water and record it. Subtract the mass of the paraffin block and the sinker when both are in water from the combined mass if the paraffin block in air and the sinker in water to get the apparent loss in the mass of the paraffin block. Divide the mass of the paraffin block in air by the apparent loss in mass of the clock to obtain its relative density. Compare the average value with the accepted value or the theoretical value and compute for the percentage error.

Answer 1

$\text{mass of bodies in water:}$

$m_{sw}=122.4g$

$m_{pw}+m_{sw}=126.5g$

$\text{mass of bodies in the air:}$

$m_{pa}=68.3g$

$m_{pa}+m_{sw} = 68.3+122.4 = 190.7g$

$\Delta m_p= (m_{pa}+m_{sw})-(m_{pw}+m_{sw})=190.7-126.5 =64.2g$

$RD = \frac{m_{pa}}{\Delta m_p}=\frac{68.3}{64.2}=1.06$

$\text{Theoretical relative density:}$

$\text{paraffin density: }\rho_p= 900\frac{kg}{m^3}$

$\text{density of water: }\rho _w =1000\frac{kg}{m^3}$

$RD_t= \frac{\rho _p}{\rho_w}= \frac{900}{1000}= 0.9$

$\text{percentage error:}$

$E_p= \frac{RD-RD_t}{RD_t}*100=\frac{1.06-0.9}{0.9}*100=17 \%$

$\text{Answer: }RD = 1.06;\text{percentage error }17 \%$